Question

calculate the following limits at infinity by dividing every term in the numerator and denominator by the highest power of ( x ) in the denominator. show all necessary work.

7.) ( limlimits_{x \to infty} \frac{x^2 + 2x + 1}{x^3 - 8} )

8.) ( limlimits_{x \to infty} \frac{2x^3 - 1}{6x^3 + 2x^2 - 4x + 3} )

9.) ( limlimits_{x \to -infty} \frac{4x^2 + 3x - 2}{7x + 1} )

10.) ( limlimits_{x \to -infty} \frac{sqrt{9x^4 + 2x^2 + 3}}{5x - 2x^2} )

11.) ( limlimits_{x \to infty} \frac{sqrt{x^6 + 8}}{4x^2 + sqrt{3x^4 + 1}} )

Explanation:

Problem 7

Step1: Identify highest denominator power

Highest power of $x$ in denominator is $x^3$.

Step2: Divide all terms by $x^3$

$$\lim_{x \to \infty} \frac{\frac{x^2}{x^3}+\frac{2x}{x^3}+\frac{1}{x^3}}{\frac{x^3}{x^3}-\frac{8}{x^3}} = \lim_{x \to \infty} \frac{\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3}}{1-\frac{8}{x^3}}$$

Step3: Evaluate limit at $\infty$

As $x \to \infty$, $\frac{1}{x},\frac{2}{x^2},\frac{1}{x^3},\frac{8}{x^3} \to 0$.
$$\frac{0+0+0}{1-0} = 0$$

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Problem 8

Step1: Identify highest denominator power

Highest power of $x$ in denominator is $x^3$.

Step2: Divide all terms by $x^3$

$$\lim_{x \to \infty} \frac{\frac{2x^3}{x^3}-\frac{1}{x^3}}{\frac{6x^3}{x^3}+\frac{2x^2}{x^3}-\frac{4x}{x^3}+\frac{3}{x^3}} = \lim_{x \to \infty} \frac{2-\frac{1}{x^3}}{6+\frac{2}{x}-\frac{4}{x^2}+\frac{3}{x^3}}$$

Step3: Evaluate limit at $\infty$

As $x \to \infty$, $\frac{1}{x^3},\frac{2}{x},\frac{4}{x^2},\frac{3}{x^3} \to 0$.
$$\frac{2-0}{6+0-0+0} = \frac{2}{6} = \frac{1}{3}$$

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Problem 9

Step1: Identify highest denominator power

Highest power of $x$ in denominator is $x$.

Step2: Divide all terms by $x$

$$\lim_{x \to -\infty} \frac{\frac{4x^2}{x}+\frac{3x}{x}-\frac{2}{x}}{\frac{7x}{x}+\frac{1}{x}} = \lim_{x \to -\infty} \frac{4x+3-\frac{2}{x}}{7+\frac{1}{x}}$$

Step3: Evaluate limit at $-\infty$

As $x \to -\infty$, $4x \to -\infty$, $\frac{2}{x},\frac{1}{x} \to 0$.
$$\frac{-\infty+3-0}{7+0} = -\infty$$

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Problem 10

Step1: Identify highest denominator power

Highest power of $x$ in denominator is $x^2$. For $\sqrt{9x^4+2x^2+3}$, $\sqrt{x^4}=|x^2|=x^2$ for all $x$.

Step2: Divide all terms by $x^2$

$$\lim_{x \to -\infty} \frac{\frac{\sqrt{9x^4+2x^2+3}}{x^2}}{\frac{5x}{x^2}-\frac{2x^2}{x^2}} = \lim_{x \to -\infty} \frac{\sqrt{\frac{9x^4}{x^4}+\frac{2x^2}{x^4}+\frac{3}{x^4}}}{\frac{5}{x}-2} = \lim_{x \to -\infty} \frac{\sqrt{9+\frac{2}{x^2}+\frac{3}{x^4}}}{\frac{5}{x}-2}$$

Step3: Evaluate limit at $-\infty$

As $x \to -\infty$, $\frac{2}{x^2},\frac{3}{x^4},\frac{5}{x} \to 0$.
$$\frac{\sqrt{9+0+0}}{0-2} = \frac{3}{-2} = -\frac{3}{2}$$

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Problem 11

Step1: Identify highest power

Highest power across numerator/denominator is $x^3$ (since $\sqrt{x^6}=|x^3|=x^3$ for $x \to \infty$).

Step2: Divide all terms by $x^3$

$$\lim_{x \to \infty} \frac{\frac{\sqrt{x^6+8}}{x^3}}{\frac{4x^2}{x^3}+\frac{\sqrt{3x^4+1}}{x^3}} = \lim_{x \to \infty} \frac{\sqrt{\frac{x^6}{x^6}+\frac{8}{x^6}}}{\frac{4}{x}+\sqrt{\frac{3x^4}{x^6}+\frac{1}{x^6}}} = \lim_{x \to \infty} \frac{\sqrt{1+\frac{8}{x^6}}}{\frac{4}{x}+\sqrt{\frac{3}{x^2}+\frac{1}{x^6}}}$$

Step3: Evaluate limit at $\infty$

As $x \to \infty$, $\frac{8}{x^6},\frac{4}{x},\frac{3}{x^2},\frac{1}{x^6} \to 0$.
$$\frac{\sqrt{1+0}}{0+\sqrt{0+0}} = \frac{1}{4+\sqrt{3}}$$

Answer:

7.) $0$
8.) $\frac{1}{3}$
9.) $-\infty$
10.) $-\frac{3}{2}$
11.) $\frac{1}{4+\sqrt{3}}$